Horsepower requirement of the tugboat towing the pipeline

Discussion in 'Boat Design' started by Ali Golshani, Nov 7, 2017.

  1. baeckmo
    Joined: Jun 2009
    Posts: 1,002
    Likes: 91, Points: 48, Legacy Rep: 1165
    Location: Sweden

    baeckmo Hydrodynamics

    Hello Ali, you asked for an example, here we go. I'll show two "basic engineering" methods as opposed to our button-pressing phantoms.......;

    The resistance comes from two parts, the surface friction and the dynamic stagnation pressure loss. Since you are dealing with pipes, we can use the pipe friction calculation for this example; somewhere, somebody has already done a friction loss estimate (and you could use their numbers directly). It is based on the wetted surface, so imagine your pipe is having an inner dia of 2.75 m. The water does not know whether it is inside or outside the wall. Let’s assume it is floating half submerged.

    • First, calculate Reynolds number: Re = V*D/ v; where V=speed, m/s, D=diameter, m, v=kinematic viscosity. Here: Re = 2.75*2/1e-6; dvs Re = 8250000.

    • Estimate surface roughness, ks. Say ks ~0.1 mm. Then D/ks = 27500.

    • In pipe friction diagram (lambda over Re, according to Nikuradse f.i.) find lambda for your Re and D/ks. In this case, lambda ~0.09.

    • Calculate pipe friction loss from pf = lambda*L/D*pdyn; where L = pipe length and pdyn is dynamic pressure, i.e. V^2*density/2. Here pdyn = 4500 Pa. The pressure loss for the fully wetted pipe is thus pf = 2121 Pa. This is the pressure differential needed to move water at 3 m/s (10.8 km/h) through the pipe, against the wall friction. The total force of this pressure over the pipe opening is: Ff = pf * Area. For a half full pipe, the force is ~2121*5.94/2, i.e. Ff =6300 N.

    • Next, calculate stagnation pressure loss as ps = pf*Cr; where Cr = loss coefficient, here ~1. The force to overcome is then ps*submerged area. Here Fs = 13365 N.

    • The total resistance is then ~20000 N.
    Now the friends of marine order will raise their voices and protest, this is a calculation of pipe flow resistance, it can’t be relied upon! So, let’s see where the differences hide then. For hull friction we use the Schoenherr method, which is based on Reynolds number using total length, and total friction area.

    • Re = 3*144/1e-6, dvs Re = 4.3e8.

    • Roughness correction for L/ks = 144000/0.1
    For those having NA litterature available, you will find the friction coefficient (Schoenherr + ATTC correction) roughly as: cf = 0.0023. This gives a friction force of Ff = 6438 N (compared to 6300 N). The stagnation force will be the same in both cases.

    So, at 3 m/s, the total force is ~20000 N/pipe. Now that speed (6 knots) is probably too high. If we use 2 m/s, there is plenty of time for one round trip per hour, including mooring and load securing. Since all forces depend on speed^2, the total “fluid” force at 2 m/s will be 8740 N/pipe. To this add current, 1 m/s and you are back to 20 kN/pipe. Then add friction to rails (as noted by Gonzo) and safety margins for wind and weather. You would probably need maximum 30 kN/pipe.

    The tug dimensioning factor is then how many pipes have to be delivered each trip. Say that 5 lengths/hour can be connected, then the tug must have a minimum towing force of 150 kN at 6 knots. The required transport power is then 150 * 3 kW.

    Propeller efficiency during towing and trawling is low, due to low advance coefficient, so in spite of what TANSL's computer model states, use 40 á 45 % as propulsion efficiency. The towing power is then ~1100 kW for this example.
     
    Ali Golshani and Barry like this.
  2. TANSL
    Joined: Sep 2011
    Posts: 4,639
    Likes: 86, Points: 58, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Interesting calculation, @baeckmo.
    The 65% efficiency that I have considered has not been decided by my computer but by myself. If it is an error, it is my error. Fortunately those things still can not be deduced by a computer, they are in the hands of the designer.
     
    Ali Golshani likes this.
  3. Ali Golshani
    Joined: Nov 2017
    Posts: 11
    Likes: 0, Points: 1
    Location: Tehran

    Ali Golshani Junior Member

    upload_2017-11-13_10-29-31.png
    Hi all, does above paragraph work for me?
     
  4. Ali Golshani
    Joined: Nov 2017
    Posts: 11
    Likes: 0, Points: 1
    Location: Tehran

    Ali Golshani Junior Member

    No, it is Tugboat. Thanks
     
  5. Ali Golshani
    Joined: Nov 2017
    Posts: 11
    Likes: 0, Points: 1
    Location: Tehran

    Ali Golshani Junior Member

    Thanks a lot.
     
  6. gonzo
    Joined: Aug 2002
    Posts: 12,406
    Likes: 208, Points: 63, Legacy Rep: 2031
    Location: Milwaukee, WI

    gonzo Senior Member



    I believe the maximum friction will be between the pipes and the steel guides.
     
  7. baeckmo
    Joined: Jun 2009
    Posts: 1,002
    Likes: 91, Points: 48, Legacy Rep: 1165
    Location: Sweden

    baeckmo Hydrodynamics

    One issue that is not dealt with yet is the downforce created by the flow around the square (plugged) pipe end forward. Neither the "basic" model nor the Holtrop Mennen regression model gives any info on this (btw I think the H+M model is limited to far lower length/beam ratios than we have in this case). To satisfy my own curiosity, I made some rude CFD-simulations with semisubmerged single, three and five pipes; the latter two in flat bundles.

    The result (at 3 m/s) shows good agreement with the "basic engineering" drag force for the single pipe; 19200 N. In addition there is a downforce of 6540 N, acting over the forward 3 to 4 meters. This phenomenon is well known among tug crews, and will limit the towing speed in order to avoid "digging in", or diving.

    The drag for bundled three is 52020 N with a downforce of 36300 N, i.e. a reduction in drag/pipe, but a substantial increase in diving force. For five pipes in a flat bundle, the drag is 88500 N and the diving force has increased to 64400 N (!). This means that it is not practical to tow bundles of five at 3 m/s (6 knot), even if the tug can deliver the towing force. Again, this force is proportional to speed^2; so I would estimate 4 knot to be about the maximum towing speed for flat bundles.

    That said, I have to keep nagging about propeller efficiency during towing. This value has a major influence over the operational conditions, safety and economy. It is not ok to pick a value from the end of the wet finger in the wind. During trawling operation at speeds around 3 to 4 knots, propeller efficiencies as low as 25 to 28 % are normal with fixed pitch propellers. Towing with CP prop in Kort nozzle and slightly higher speed may result in 40-ish percentages. Stating 65% efficiency at towing is misleading and unrealistic. Sometimes we have to "guesstimate" in engineering, but in this case we have accurate knowledge to fall back upon when it comes to propulsion.
     
    Ali Golshani likes this.
  8. Ad Hoc
    Joined: Oct 2008
    Posts: 5,350
    Likes: 207, Points: 63, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    Indeed :)

    How true...and based upon "sound" engineering :p

    Dont forget, there will be a small amount of residuary resistance to add too. At these low Fns, merely adding 10% to the total, should suffice.
     
  9. Ali Golshani
    Joined: Nov 2017
    Posts: 11
    Likes: 0, Points: 1
    Location: Tehran

    Ali Golshani Junior Member

    Hi and thanks for the reply.
    Due to existing balance concrete ballasts, the drag force between concrete ballast and water in towing operation should be calculated and added to the calculations. I attached the drawing of pipeline+ ballast section. The ballast are every 45m. Can we consider a modified drag coefficient for pipeline to consider the effect of ballasts indirectly ?
    Thanks



     

    Attached Files:

  10. Squidly-Diddly
    Joined: Sep 2007
    Posts: 1,217
    Likes: 27, Points: 48, Legacy Rep: 304
    Location: SF bay

    Squidly-Diddly Senior Member

    When he says "add them, not subtract" he means you might be "adding" vectors going in different directions, which most people would say is "subtracting".

    Its sort of like accounting. You "add" both the positive deposits and negative withdraws together.

    However, 10 tons pushing one way and 10 tons pushing the other way might add up to zero but its not the same as "no force involved", just like an account with a million dollar deposit and million dollar withdrawal ain't quite the same as "no activity".
     

  11. TANSL
    Joined: Sep 2011
    Posts: 4,639
    Likes: 86, Points: 58, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Yes, forces are vectors and then "adding forces" means "adding vectors".
    If you subtract a negative scalar value from a set of scalars, you are probably adding it. English semantics is not my strong point.
    About banks account I'm afraid not to know much.
     
Loading...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.